Statistics for Management and Economics Abbreviated 10th Edition by Gerald Keller - Test Bank

Statistics for Management and Economics Abbreviated 10th Edition by Gerald Keller – Test Bank

 

Instant Download – Complete Test Bank With Answers

 

 

Sample Questions Are Posted Below

 

CHAPTER 8: CONTINUOUS PROBABILITY DISTRIBUTIONS

 

TRUE/FALSE

 

1. Since there is an infinite number of values a continuous random variable can assume, the probability of each individual value is virtually 0.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

2. A continuous probability distribution represents a random variable having an infinite number of outcomes which may assume any number of values within an interval.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

3. Continuous probability distributions describe probabilities associated with random variables that are able to assume any finite number of values along an interval.

 

ANS:  F                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

4. A continuous random variable X has a uniform distribution between 10 and 20 (inclusive), then the probability that X falls between 12 and 15 is 0.30.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

5. A continuous random variable is one that can assume an uncountable number of values.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

6. A continuous random variable X has a uniform distribution between 5 and 15 (inclusive), then the probability that X falls between 10 and 20 is 1.0.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

7. A continuous random variable X has a uniform distribution between 5 and 25 (inclusive), then P(X = 15) = 0.05.

 

ANS:  F                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

8. We distinguish between discrete and continuous random variables by noting whether the number of possible values is countable or uncountable.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

9. In practice, we frequently use a continuous distribution to approximate a discrete one when the number of values the variable can assume is countable but large.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

10. Let X represent weekly income expressed in dollars. Since there is no set upper limit, we cannot identify (and thus cannot count) all the possible values. Consequently, weekly income is regarded as a continuous random variable.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

11. To be a legitimate probability density function, all possible values of f(x) must be non-negative.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

12. To be a legitimate probability density function, all possible values of f(x) must lie between 0 and 1 (inclusive).

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

13. The sum of all values of f(x) over the range of [a, b] must equal one.

 

ANS:  F                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

14. A probability density function shows the probability for each value of X.

 

ANS:  F                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

15. If X is a continuous random variable on the interval [0, 10], then P(X > 5) = P(X ³ 5).

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

16. If X is a continuous random variable on the interval [0, 10], then P(X = 5) = f(5) = 1/10.

 

ANS:  F                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

17. If a point y lies outside the range of the possible values of a random variable X, then f(y) must equal zero.

 

ANS:  T                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

MULTIPLE CHOICE

 

1. Which of the following is always true for all probability density functions of continuous random variables?

a.	The probability at any single point is zero.
b.	They contain an uncountable number of possible values.
c.	The total area under the density function f(x) equals 1.
d.	All of these choices are true.

 

 

ANS:  D                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

2. The probability density function, f(x), for any continuous random variable X, represents:

a.	all possible values that X will assume within some interval a £ x £ b.
b.	the probability that X takes on a specific value x.
c.	the height of the density function at x.
d.	None of these choices.

 

 

ANS:  C                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

3. Which of the following represents a difference between continuous and discrete random variables?

a.	Continuous random variables assume an uncountable number of values, and discrete random variables do not.
b.	The probability for any individual value of a continuous random variable is zero, but for discrete random variables it is not.
c.	Probability for continuous random variables means finding the area under a curve, while for discrete random variables it means summing individual probabilities.
d.	All of these choices are true.

 

 

ANS:  D                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

4. Suppose f(x) = 0.25. What range of possible values can X take on and still have the density function be legitimate?

a.	[0, 4]	c.	[-2, +2]
b.	[4, 8]	d.	All of these choices are true.

 

 

ANS:  D                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

5. What is the shape of the probability density function for a uniform random variable on the interval [a, b]?

a.	A rectangle whose X values go from a to b.
b.	A straight line whose height is 1/(b – a) over the range [a, b].
c.	A continuous probability density function with the same value of f(x) from a to b.
d.	All of these choices are true.

 

 

ANS:  D                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

6. Which of the following is true about f(x) when X has a uniform distribution over the interval [a, b]?

a.	The values of f(x) are different for various values of the random variable X.
b.	f(x) equals one for each possible value of X.
c.	f(x) equals one divided by the length of the interval from a to b.
d.	None of these choices.

 

 

ANS:  C                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

7. Suppose f(x) = 1/4 over the range a £ x £ b, and suppose P(X > 4) = 1/2. What are the values for a and b?

a.	0 and 4
b.	2 and 6
c.	Can be any range of x values whose length (b – a) equals 4.
d.	Cannot answer with the information given.

 

 

ANS:  B                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

8. The probability density function f(x) for a uniform random variable X defined over the interval [2, 10] is

a.	0.20	c.	4
b.	8	d.	None of these choices.

 

 

ANS:  D                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

9. If the random variable X has a uniform distribution between 40 and 50, then P(35 £ X £ 45) is:

a.	1.0	c.	0.1
b.	0.5	d.	undefined.

 

 

ANS:  B                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

10. The probability density function f(x) of a random variable X that has a uniform distribution between a and b is

a.	(b + a)/2	c.	(a – b)/2
b.	1/b – 1/a	d.	None of these choices.

 

 

ANS:  D                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

11. Which of the following does not represent a continuous uniform random variable?

a.	f(x) = 1/2 for x between -1 and 1, inclusive.
b.	f(x) = 10 for x between 0 and 1/10, inclusive.
c.	f(x) = 1/3 for x = 4, 5, 6.
d.	None of these choices represents a continuous uniform random variable.

 

 

ANS:  C                    PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

COMPLETION

 

1. A(n) ____________________ random variable is one that assumes an uncountable number of possible values.

 

ANS:  continuous

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

2. For a continuous random variable, the probability for each individual value of X is ____________________.

 

ANS:

zero

0

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

3. Probability for continuous random variables is found by finding the ____________________ under a curve.

 

ANS:  area

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

4. A(n) ____________________ random variable has a density function that looks like a rectangle and you can use areas of a rectangle to find probabilities for it.

 

ANS:  uniform

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

5. Suppose X is a continuous random variable for X between a and b. Then its probability ____________________ function must non-negative for all values of X between a and b.

 

ANS:  density

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

6. The total area under f(x) for a continuous random variable must equal ____________________.

 

ANS:

1

one

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

7. The probability density function of a uniform random variable on the interval [0, 5] must be ____________________ for 0 £ x £ 5.

 

ANS:

1/5

0.20

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

8. To find the probability for a uniform random variable you take the ____________________ times the ____________________ of its corresponding rectangle.

 

ANS:

base; height

height; base

length; width

width; length

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

9. You can use a continuous random variable to ____________________ a discrete random variable that takes on a countable, but very large, number of possible values.

 

ANS:  approximate

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

SHORT ANSWER

 

1. A continuous random variable X has the following probability density function:

f(x) = 1/4, 0 £ x £ 4

Find the following probabilities:

 

a.	P(X £ 1)
b.	P(X ³ 2)
c.	P(1 £ X £ 2)
d.	P(X = 3)

 

 

ANS:

 

a.	0.25
b.	0.50
c.	0.25
d.	0

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Waiting Time

 

The length of time patients must wait to see a doctor at an emergency room in a large hospital has a uniform distribution between 40 minutes and 3 hours.

 

2. {Waiting Time Narrative} What is the probability density function for this uniform distribution?

 

ANS:

f(x) = 1/140, 40 £ x £ 180 (minutes)

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

3. {Waiting Time Narrative} What is the probability that a patient would have to wait between one and two hours?

 

ANS:

0.43

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

4. {Waiting Time Narrative} What is the probability that a patient would have to wait exactly one hour?

 

ANS:

0

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

5. {Waiting Time Narrative} What is the probability that a patient would have to wait no more than one hour?

 

ANS:

0.143

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

6. The time required to complete a particular assembly operation has a uniform distribution between 25 and 50 minutes.

 

a.	What is the probability density function for this uniform distribution?
b.	What is the probability that the assembly operation will require more than 40 minutes to complete?
c.	Suppose more time was allowed to complete the operation, and the values of X were extended to the range from 25 to 60 minutes. What would f(x) be in this case?

 

 

ANS:

 

a.	f(x) = 1/25, 25 £ x £ 50
b.	0.40
c.	f(x) = 1/35, 25 £ x £ 60

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

7. Suppose f(x) equals 1/50 on the interval [0, 50].

 

a.	What is the distribution of X?
b.	What does the graph of f(x) look like?
c.	Find P(X £ 25)
d.	Find P(X ³ 25)
e.	Find P(X = 25)
f.	Find P(0 < X < 3)
g.	Find P(-3 < X < 0)
h.	Find P(0 < X < 50)

 

 

ANS:

 

a.	X has a uniform distribution on the interval [0, 50].
b.	f(x) forms a rectangle of height 1/50 from x = 0 to x = 50.
c.	0.50
d.	0.50
e.	0
f.	0.06
g.	0.06
h.	1.00

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Electronics Test

 

The time it takes a student to finish a electronics test has a uniform distribution between 50 and 70 minutes.

 

8. {Electronics Test Narrative} What is the probability density function for this uniform distribution?

 

ANS:

f(x) = 1/20, 50 £ x £ 70

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

9. {Electronics Test Narrative} Find the probability that a student will take more than 60 minutes to finish the test.

 

ANS:

0.50

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

10. {Electronics Test Narrative} Find the probability that a student will take no less than 55 minutes to finish the test.

 

ANS:

0.75

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

11. {Electronics Test Narrative} Find the probability that a student will take exactly one hour to finish the test.

 

ANS:

0

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

12. {Electronics Test Narrative} What is the median amount of time it takes a student to finish the test?

 

ANS:

60 minutes

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

13. {Electronics Test Narrative} What is the mean amount of time it takes a student to finish the test?

 

ANS:

60 minutes

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Subway Waiting Time

 

At a subway station the waiting time for a subway is found to be uniformly distributed between 1 and 5 minutes.

 

14. {Subway Waiting Time Narrative} What is the probability density function for this uniform distribution?

 

ANS:

f(x) = 1/4, 1 £ x £ 5

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

15. {Subway Waiting Time Narrative} What is the probability of waiting no more than 3 minutes?

 

ANS:

0.50

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

16. {Subway Waiting Time Narrative} What is the probability that the subway arrives in the first minute and a half?

 

ANS:

0.125

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

17. {Subway Waiting Time Narrative} What is the median waiting time for this subway?

 

ANS:

3 minutes

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.01

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

18. A national standardized testing company can tell you your relative standing on an exam without divulging the mean or the standard deviation of the exam scores.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

19. If your golf score is 3 standard deviations below the mean, its corresponding value on the Z distribution is -3.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

20. If we standardize the normal curve, we express the original X values in terms of their number of standard deviations away from the mean.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

21. A normal distribution is symmetric; therefore the probability of being below the mean is 0.50 and the probability of being above the mean is 0.50.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

22. A random variable X is standardized by subtracting the mean and dividing by the variance.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

23. A random variable X has a normal distribution with mean 132 and variance 36. If x = 120, its corresponding value of Z is 2.0.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

24. A random variable X has a normal distribution with a mean of 250 and a standard deviation of 50. Given that X = 175, its corresponding value of Z is -1.50.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

25. Given that Z is a standard normal random variable, a negative value of Z indicates that the standard deviation of Z is negative.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

26. In the standard normal distribution, z_0.05 = 1.645 means that 5% of all values of z are below 1.645 and 95% are above it.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

27. The probability that a standard normal random variable Z is less than -3.5 is approximately 0.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

28. If the value of Z is z = 99, that means you are at the 99th percentile on the Z distribution.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

29. The 10th percentile of a Z distribution has 10% of the Z-values lying above it.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

30. The probability that Z is less than -2 is the same as one minus the probability that Z is greater than +2.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

31. Suppose X has a normal distribution with mean 70 and standard deviation 5. The 50th percentile of X is 70.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

32. Which of the following is not a characteristic for a normal distribution?

a.	It is symmetrical.
b.	The mean is always zero.
c.	The mean, median, and mode are all equal.
d.	It is a bell-shaped distribution.

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

33. If X has a normal distribution with mean 60 and standard deviation 6, which value of X corresponds with the value z = 1.96?

a.	x = 71.76
b.	x = 67.96
c.	x = 61.96
d.	x = 48.24

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

34. A standard normal distribution is a normal distribution with:

a.	a mean of zero and a standard deviation of one.
b.	a mean of one and a standard deviation of zero.
c.	a mean always larger than the standard deviation.
d.	None of these choices.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

35. What proportion of the data from a normal distribution is within two standard deviations from the mean?

a.	0.3413
b.	0.4772
c.	0.6826
d.	0.9544

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

36. Given that Z is a standard normal random variable, the area to the left of a value z is expressed as

a.	P(Z ³ z)
b.	P(Z £ z)
c.	P(0 £ Z £ z)
d.	P(Z ³ –z)

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

37. Given that Z is a standard normal variable, the variance of Z:

a.	is always greater than 2.0.
b.	is always greater than 1.0.
c.	is always equal to 1.0.
d.	cannot assume a specific value.

 

 

ANS:

C

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

38. Given that Z is a standard normal random variable, a negative value (z) on its distribution would indicate:

a.	z is to the left of the mean.
b.	the standard deviation of this Z distribution is negative.
c.	the area between zero and the value z is negative.
d.	None of these choices.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

39. A larger standard deviation of a normal distribution indicates that the distribution becomes:

a.	narrower and more peaked.
b.	flatter and wider.
c.	more skewed to the right.
d.	more skewed to the left.

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

40. In its standardized form, the normal distribution:

a.	has a mean of 0 and a standard deviation of 1.
b.	has a mean of 1 and a variance of 0.
c.	has an area equal to 0.5.
d.	cannot be used to approximate discrete probability distributions.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

41. Most values of a standard normal distribution lie between:

a.	0 and 1
b.	-3 and 3
c.	0 and 3
d.	minus infinity and plus infinity

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

42. Stacy took a math test whose mean was 70 and standard deviation was 5. The total points possible was 100. Stacey’s results were reported to be at the 95th percentile. What was Stacey’s actual exam score, rounded to the nearest whole number?

a.	95
b.	78
c.	75
d.	62

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

43. Tanner took a statistics test whose mean was 80 and standard deviation was 5. The total points possible was 100. Tanner’s score was 2 standard deviations below the mean. What was Tanner’s score, rounded to the nearest whole number?

a.	78
b.	70
c.	90
d.	None of these choices.

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

44. Lamont took a psychology exam whose mean was 70 with standard deviation 5. He also took a calculus exam whose mean was 80 with standard deviation 10. He scored 85 on both exams. On which exam did he do better compared to the other students who took the exam?

a.	He did better on the psychology exam, comparatively speaking.
b.	He did better on the calculus exam, comparatively speaking.
c.	He did the same on both exams, relatively speaking.
d.	Cannot tell without more information.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

45. Suppose Lamont’s exam score was at the 80^th percentile on an exam whose mean was 90. What was Lamont’s exam score?

a.	76.81
b.	72.00
c.	80.00
d.	Cannot tell without more information.

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

46. Suppose X has a normal distribution with mean 40 and standard deviation 2. Shifting all the X values to the right 10 units results in a normal distribution with mean ____________________ and standard deviation ____________________.

 

ANS:

50; 2

fifty; two

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

47. ____________________ the value of s in a normal distribution will make it wider.

 

ANS:

Increasing

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

48. We standardize a random variable by subtracting its ____________________ and dividing by its ____________________.

 

ANS:

mean; standard deviation

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

49. Suppose X has a normal distribution with mean 10 and standard deviation 2. The probability that X is less than 8 is equal to the probability that Z is less than ____________________.

 

ANS:

-1

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

50. P(Z > 1.9) = ____________________ P(Z < 1.9).

 

ANS:

1 –

1-

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

51. P(1 < Z < 2) = P(Z < 2) – ____________________.

 

ANS:

P(Z < 1)

P(Z<1)

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

52. The mean of the standard normal distribution is ____________________ and the standard deviation is ____________________.

 

ANS:

0; 1

zero; one

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

53. P(Z > 3.00) is approximately ____________________.

 

ANS:

0

zero

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

54. P(Z < 3.00) is approximately ____________________.

 

ANS:

1

one

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

55. Suppose X is a normal random variable with mean 70 and standard deviation 3. Then P(X = 3) = ____________________.

 

ANS:

0

zero

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

56. Z_.025 is the value of Z such that the area to the ____________________ of Z is .9750.

 

ANS:

left

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

Battery Life

 

A certain brand of batteries has a lifetime that has a normal distribution with a mean of 3,750 hours and a standard deviation of 300 hours.

 

57. {Battery Life Narrative} What proportion of these batteries will last for more than 4,000 hours?

 

ANS:

0.2033

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

58. {Battery Life Narrative} What proportion of these batteries will last less than 3,600 hours?

 

ANS:

0.3085

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

59. {Battery Life Narrative} What proportion of these batteries will last between 3,800 and 4,100 hours?

 

ANS:

0.3115

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

60. {Battery Life Narrative} What lifetime should the manufacturer advertise for these batteries in order that only 2% of the lamps will wear out before the advertised lifetime?

 

ANS:

3135

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Diet

 

Researchers studying the effects of a new diet found that the weight loss over a one-month period by those on the diet was normally distributed with a mean of 10 pounds and a standard deviation of 5 pounds.

 

61. {Diet Narrative} What proportion of the dieters lost more than 12 pounds?

 

ANS:

0.3446

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

62. {Diet Narrative} What proportion of the dieters gained weight?

 

ANS:

0.0228

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

63. {Diet Narrative} If a dieter is selected at random, what is the probability that the dieter lost more than 7.5 pounds?

 

ANS:

0.6915

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

64. Let X be a normally distributed random variable with a mean of 12 and a standard deviation of 1.5. What proportions of the values of X are:

 

a.	less than 14
b.	more than 8
c.	between 10 and 13

 

 

ANS:

 

a.	0.9082
b.	0.9962
c.	0.6568

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

65. If Z is a standard normal random variable, find the value z for which:

 

a.	the area between 0 and z is 0.3729
b.	the area to the right of z is 0.7123
c.	the area to the left of z is 0.1736
d.	the area between –z and z is 0.6630

 

 

ANS:

 

a.	1.14
b.	-.56
c.	-.94
d.	0.96

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

66. If Z is a standard normal random variable, find the following probabilities:

 

a.	P(Z £ -1.77)
b.	P(Z ³ -1.96)
c.	P(0.35 £ Z £ 0.85)
d.	P(-2.88 £ Z £ -2.15)
e.	P(Z £ 1.45)

 

 

ANS:

 

a.	0.0384
b.	0.9750
c.	0.1655
d.	0.0138
e.	0.9265

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Calculus Scores

 

Scores of high school students on a national calculus exam were normally distributed with a mean of 86 and a standard deviation of 4. (Total possible points = 100.)

 

67. {Calculus Scores Narrative} What is the probability that a randomly selected student will have a score of 80 or higher?

 

ANS:

0.9332

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

68. {Calculus Scores Narrative} What is the probability that a randomly selected student will have a score between 80 and 90?

 

ANS:

0.7745

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

69. {Calculus Scores Narrative} What is the probability that a randomly selected student will have a score of 94 or lower?

 

ANS:

0.9772

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Checking Accounts

 

A bank has determined that the monthly balances of the checking accounts of its customers are normally distributed with an average balance of $1,200 and a standard deviation of $250.

 

70. {Checking Accounts Narrative} What proportion of customers have monthly balances less than $1,000?

 

ANS:

0.2119

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

71. {Checking Accounts Narrative} What proportion of customers have monthly balances more than $1,125?

 

ANS:

0.6179

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

72. {Checking Accounts Narrative} What proportion of customers have monthly balances between $950 and $1,075?

 

ANS:

0.1498

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

IT Graduates Salary

 

The recent average starting salary for new college graduates in IT systems is $47,500. Assume salaries are normally distributed with a standard deviation of $4,500.

 

73. {IT Graduates Salary Narrative} What is the probability of a new graduate receiving a salary between $45,000 and $50,000?

 

ANS:

0.4246

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

74. {IT Graduates Salary Narrative} What is the probability of a new graduate getting a starting salary in excess of $55,000?

 

ANS:

0.0475

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

75. {IT Graduates Salary Narrative} What percent of starting salaries are no more than $42,250?

 

ANS:

12.10%

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

76. {IT Graduates Salary Narrative} What is the cutoff for the bottom 5% of the salaries?

 

ANS:

$40,097.50

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

77. {IT Graduates Salary Narrative} What is the cutoff for the top 3% of the salaries?

 

ANS:

$55,960

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

78. A worker earns $16 per hour at a plant and is told that only 5% of all workers make a higher wage. If the wage is assumed to be normally distributed and the standard deviation of wage rates is $5 per hour, find the average wage for the plant workers per hour.

 

ANS:

P(X > 16) = .05 Þ (16 – m) / 5 = 1.645 Þ m = $7.78

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.02

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

79. The mean and the variance of an exponential distribution are equal to each other.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

80. The exponential distribution is suitable to model the length of time that elapses before the first telephone call is received by a switchboard.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

81. The mean and standard deviation of an exponential random variable are equal to each other.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

82. In the exponential distribution, X takes on an infinite number of possible values in the given range.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

83. If the mean of an exponential distribution is 2, then the value of the parameter l is 2.0.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

84. If the random variable X is exponentially distributed and the parameter of the distribution l = 4, then P(X £ 1) = 0.25.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

85. If the random variable X is exponentially distributed with parameter l = 5, then the variance of X, s² = V(X) = 0.04.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

86. If the random variable X is exponentially distributed with parameter l = 0.05, then the variance of X, s² = V(X) = 20.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

87. If the random variable X is exponentially distributed with parameter l = 0.05, then the probability P(X > 20) = 0.3679.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

88. If the random variable X is exponentially distributed with parameter l = 0.05, then the probability P(X < 5) = .2865.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

89. If the random variable X is exponentially distributed with parameter l = 2, then the probability that X is between 1 and 2 equals the probability that X is between 2 and 3.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

90. Which of the following is true for an exponential distribution with parameter l?

a.	m = 1/l
b.	s = 1/l
c.	The Y-intercept of f(x) is l.
d.	All of these choices are true.

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

91. If the random variable X is exponentially distributed with parameter l = 3, then the probability P(X ³ 2) equals:

a.	0.3333
b.	0.5000
c.	0.6667
d.	0.0025

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

92. If the random variable X is exponentially distributed with parameter l = 1.5, then the probability P(2 £ X £ 4), up to 4 decimal places, is

a.	0.6667
b.	0.0473
c.	0.5000
d.	0.2500

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

93. If the random variable X is exponentially distributed with parameter l = 4, then the probability P(X £ 0.25), up to 4 decimal places, is

a.	0.6321
b.	0.3679
c.	0.2500
d.	None of these choices.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

94. Which of the following can have an exponential distribution?

a.	Time between phone calls coming in to a technical support desk.
b.	Time until the first customer arrives at the bank in the morning.
c.	Lifetime of a new battery.
d.	All of these choices are true.

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

95. The exponential density function f(x):

a.	is bell-shaped.
b.	is symmetrical.
c.	approaches infinity as x approaches zero.
d.	approaches zero as x approaches infinity.

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

96. If the random variable X is exponentially distributed, then the mean of X will be:

a.	greater than the median.
b.	less than the median.
c.	equal to the median.
d.	Cannot tell; the answer depends on what l is.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

97. If the mean of an exponential distribution is 2, then the value of the parameter l is

a.	0
b.	2.0
c.	0.5
d.	1.0

 

 

ANS:

C

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

98. If the parameter of an exponential distribution is 1, then which of the following is true?

a.	The density function is e–x for x ³ 0.
b.	The mean is equal to 1.
c.	The standard deviation and variance are both equal to 1.
d.	All of these choices are true.

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

99. A random variable with density function e^–x for x ³ 0 has an exponential distribution with l = ____________________.

 

ANS:

one

1

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

100. A random variable with density function e^–x for x ³ 0 has an exponential distribution whose mean is ____________________.

 

ANS:

one

1

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

101. A random variable with density function 0.01e^–x/100 for x ³ 0 has an exponential distribution whose mean is ____________________.

 

ANS:

100

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

102. The shape of the density function for an exponential distribution is ____________________.

 

ANS:

skewed

positively skewed

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

103. The mean of an exponential random variable is ____________________ the median.

 

ANS:

greater than

>

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

104. If X has an exponential distribution, the possible values of X are from ____________________ to infinity.

 

ANS:

zero

0

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

105. An exponential random variable is an example of a(n) ____________________ random variable.

 

ANS:

continuous

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

106. If X has an exponential distribution with parameter l, then f(0) = ____________________.

 

ANS:

l

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

107. If X has an exponential distribution with parameter l, then the mean of X is ______________.

 

ANS:

1/l

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

108. If X has an exponential distribution, then f(x) approaches ____________________ as x approaches infinity.

 

ANS:

zero

0

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

109. The y-intercept of the density function for an exponential distribution with parameter 10 is ____________________.

 

ANS:

10

(0, 10)

y = 10

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

110. If X has an exponential distribution, its ____________________ is equal to its ____________________.

 

ANS:

mean; standard deviation

standard deviation; mean

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

111. Let X be an exponential random variable with l = 1.50. Find the following:

 

a.	P(X ³ 2)
b.	P(X £ 4)
c.	P(1 £ X £ 3)
d.	P(X = 1)

 

 

ANS:

 

a.	0.0498 (note that f(x) = 1.50e-1.50x for x ³ 0)
b.	0.9975
c.	0.2120
d.	0

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

112. Let X be an exponential random variable with l = 1.50. Find the following:

 

a.	f(x)
b.	The y-intercept of f(x)

 

 

ANS:

 

a.	f(x) = 1.50e-1.50x for x ³ 0
b.	(0, 1.50)

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

113. Suppose X has an exponential distribution with mean 2. Find f(x).

 

ANS:

f(x) = 0.50e^-0.50x for x ³ 0

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Truck Salesman

 

A used truck salesman in a small town states that, on the average, it takes him 5 days to sell a truck. Assume that the probability distribution of the length of time between sales is exponentially distributed.

 

114. {Truck Salesman Narrative} What is the probability that he will have to wait at least 8 days before making another sale?

 

ANS:

0.2019 (Note l is 1/5 = 0.20 days.)

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

115. {Truck Salesman Narrative} What is the probability that he will have to wait between 6 and 10 days before making another sale?

 

ANS:

0.1659

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Repair Time

 

The time it takes a technician to fix a telephone problem is exponentially distributed with a mean of 15 minutes.

 

116. {Repair Time Narrative} What is the probability density function for the time it takes a technician to fix a telephone problem?

 

ANS:

f(x) = (1/15)e^–x/15, x ³ 0

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

117. {Repair Time Narrative} What is the probability that it will take a technician less than 10 minutes to fix a telephone problem?

 

ANS:

0.4866

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

118. {Repair Time Narrative} What is the variance of the time it takes a technician to fix a telephone problem?

 

ANS:

225

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

119. {Repair Time Narrative} What is the probability that it will take a technician between 10 to 15 minutes to fix a telephone problem?

 

ANS:

0.1455

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Light Bulb Lifetime

 

The lifetime of a light bulb (in hours) is exponentially distributed with l = 0.008.

 

120. {Light Bulb Lifetime Narrative} What is the mean and standard deviation of the light bulb’s lifetime?

 

ANS:

m = s = 1/l = 1/0.008 = 125 hours

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

121. {Light Bulb Lifetime Narrative} Find the probability that a light bulb will last between 120 and 140 hours.

 

ANS:

P(120 £ X £ 140) = e^{–0.008(120)} – e^{–0.008(140)} = 0.3829 – 0.3263 = 0.0566

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

122. {Light Bulb Lifetime Narrative} Find the probability that a light bulb will last for:

 

a.	more than 125 hours.
b.	at most 125 hours.
c.	no more than 125 hours.
d.	exactly 125 hours.
e.	less than 125 hours.
f.	at least 125 hours.
g.	no less than 125 hours.

 

 

ANS:

 

a.	P(X > 125) = 0.3679
b.	P(X £ 125) = 0.6321
c.	P(X £ 125) = 0.6321
d.	P(X = 125) = 0
e.	P(X < 125) = 0.6321
f.	P(X ³ 125) = 0.3679
g.	P(X ³ 125) = 0.3679

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Counter Sales

 

Suppose that customers arrive at a counter at an average rate of three customers per minute and that their arrivals follow the Poisson model.

 

123. {Counter Sales Narrative} Write the probability density function of the distribution of the time that will elapse before the next customer arrives.

 

ANS:

Let T = Elapsed time before the next customer arrives. The random variable T follows an exponential distribution where l = 3; with mean 1/3 minute between customers. Then the probability density function of T is f(t) = 3e^-3t, t ³ 0 minutes.

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

124. {Counter Sales Narrative} Use the appropriate exponential distribution to find the probability that the next customer will arrive within 1.5 minutes.

 

ANS:

0.9889

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

125. {Counter Sales Narrative} Use the appropriate exponential distribution to find the probability that the next customer will not arrive within the next 2 minutes.

 

ANS:

0.0025

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

Phone Orders

 

The L. L. Bean catalog department that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes.

 

126. {Phone Orders Narrative} What is the value of l, the parameter of the exponential distribution in this situation?

 

ANS:

Since m = 3, then .

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

127. {Phone Orders Narrative} What proportion of customers having to hold more than 1.5 minutes will hang up before placing an order?

 

ANS:

P(X > 1.5) = e^–0.5 = 0.6065

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

128. {Phone Orders Narrative} Find the waiting time at which only 10% of the customers will continue to hold.

 

ANS:

P(X > x) = e^–lx Þ e^–x/3 = .10 Þ x = 6.908 minutes.

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

129. {Phone Orders Narrative} Find the time at which 50% of the customers will continue to hold?

 

ANS:

P(X > x) = e^–lx Þ e^–x/3 = .50 Þ x = 2.079 minutes.

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

130. {Phone Orders Narrative} What proportion of callers are put on hold longer than 3 minutes?

 

ANS:

P(X > 3) = e^–3/3 = e^–1 = 0.3679.

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

131. {Phone Orders Narrative} What is the probability that a randomly selected caller is placed on hold for fewer than 6 minutes?

 

ANS:

P(X < 6) = 1 – e^–6/3 = 1 – e^–2 = 0.8647.

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

132. {Phone Orders Narrative} What is the probability that a randomly selected caller is placed on hold for 3 to 6 minutes?

 

ANS:

P(3 < X < 6) = e^–3/3 – e^–6/3 = e^–1 – e^–2 = 0.2325.

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.03

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

133. Like that of the Student t distribution, the shape of the chi-squared distribution depends on its number of degrees of freedom.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

134. The value of c² with v degrees of freedom such that the area to its right under the chi-squared curve is equal to A is denoted by , while denotes the value such that the area to its left is A.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

135. The variance of a Student t random variable with v degrees of freedom (v > 2) is always greater than 1.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

136. We define as the value of the F with v₁ and v₂ degrees of freedom such that the area to its right under the F curve is A, while  is defined as the value such that the area to its left is A.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

137. The value of A such that P(-A £ t £ A) = 0.95, where the degrees of freedom are 20, is 2.086.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

138. The variance of a c² distribution is twice the value of its mean.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

139. As the degrees of freedom approach infinity, the values of a Student t distribution approach those of a standard normal distribution.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

140. The value of an F distribution with v₁ = 5 and v₂ = 10 degrees of freedom such that the area to its left is 0.95 is 3.33.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

141. The expected value of the Student t distribution is zero.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

142. The variance of a Student t distribution approaches zero as the degrees of freedom approaches infinity.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

143. The value of an F distribution with v₁ = 6 and v₂ = 9 degrees of freedom such that the area to its right is 0.05 is 3.37.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

144. The value of a chi-squared distribution with 5 degrees of freedom such that the area to its left is 0.10 is 1.61.

 

ANS:

T

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

145. The Student t distribution looks similar in shape to a standard normal distribution, except it is not as widely spread.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

146. The value of a chi-squared distribution with 8 degrees of freedom such that the area to its left is 0.95 is 2.73.

 

ANS:

F

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

147. The Student t distribution:

a.	is symmetrical.
b.	approaches the normal distribution as the degrees of freedom increase.
c.	has more area in the tails than the standard normal distribution does.
d.	All of these choices are true.

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

148. The Student t distribution with parameter v = 2 has a mean E(t) equal to:

a.	0
b.	1
c.	2
d.	None of these choices.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

149. The Student t distribution with parameter v = 4 has a variance V(t) equal to:

a.	4
b.	0
c.	2
d.	1

 

 

ANS:

C

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

150. Which of the following statements is correct regarding the percentile points of the F distribution?

a.	F0.10,10,20 = 1/F0.90,10,20
b.	F0.90,10,20 = 1/F0.10,20,10
c.	F0.90,10,20 = 1/F0.90,20,10
d.	F0.10,10,20 = 1/F0.10,20,10

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

151. Which of the following statements is true?

a.	The chi-squared distribution is positively skewed.
b.	All the values of the chi-squared distribution are non-negative.
c.	The shape of the chi-squared distribution depends on its degrees of freedom.
d.	All of these choices are true.

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

152. What number corresponds to t_0.05,10?

a.	1.812
b.	1.372
c.	2.228
d.	1.833

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

153. If P(t > t_.01,v) = 2.50, then the number of degrees of freedom v is:

a.	20
b.	21
c.	22
d.	23

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

154. What number corresponds to ?

a.	28.30
b.	26.22
c.	21.00
d.	5.23

 

 

ANS:

C

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

155. If , then the number of degrees of freedom v is:

a.	5
b.	6
c.	7
d.	8

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

156. What number corresponds to F_0.95,4,8?

a.	6.040
b.	3.840
c.	0.260
d.	0.166

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

157. What number corresponds to F_.025,3,5?

a.	14.88
b.	7.76
c.	12.06
d.	5.41

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

158. Suppose X has a chi-squared distribution with 10 degrees of freedom. The mean of X is:

a.	10
b.	9
c.	20
d.	None of these choices.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

159. Suppose X has a chi-squared distribution with 10 degrees of freedom. The variance of X is:

a.	20
b.	10
c.	9
d.	None of these choices.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

160. The number of parameters for an F distribution is:

a.	1
b.	2
c.	0
d.	None of these choices.

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

161. Suppose X has an F distribution. Which of the following is true?

a.	f(x) is symmetrical.
b.	All the values of X are non-negative.
c.	The mean of X is zero.
d.	All of these choices.

 

 

ANS:

B

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

162. Which of the following distributions can take on negative values?

a.	Student t
b.	c2
c.	F
d.	None of these choices.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

163. Which of the following distributions is not skewed?

a.	Student t
b.	c2
c.	F
d.	All of these distributions are skewed.

 

 

ANS:

A

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

164. Which of the following has a mean and variance that depend on degrees of freedom?

a.	Student t
b.	c2
c.	F
d.	All of these choices are true.

 

 

ANS:

D

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

165. The shape of the ____________________ distribution is similar to a normal distribution, except it has more area in the tails.

 

ANS:

Student t

t

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

166. The mean of a Student t distribution is ____________________.

 

ANS:

zero

0

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

167. The variance of a Student t distribution approaches ____________________ as the degrees of freedom increase to infinity.

 

ANS:

one

1

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

168. A c² distribution with 5 degrees of freedom has a mean of ____________________ and a variance of ____________________.

 

ANS:

5; 10

five; ten

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

169. The mean and variance of a c² distribution approach ____________________ as the degrees of freedom increase.

 

ANS:

infinity

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

170. For values of degrees of freedom greater than 100, the c² distribution can be approximated by a(n) ____________________ distribution.

 

ANS:

normal

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

171. The shape of a c² distribution is ____________________.

 

ANS:

positively skewed

skewed

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

172. The F distribution has two parameters called degrees of freedom, n₁ and n₂. We call n₁ the ____________________ degrees of freedom, and we call n₂ the ____________________ degrees of freedom.

 

ANS:

numerator; denominator

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Knowledge

 

173. As the ____________________ degrees of freedom increase, the mean of the F distribution approaches 1.

 

ANS:

denominator

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

174. The shape of an F distribution is ____________________.

 

ANS:

positively skewed

skewed

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

175. Use the t-table to find the following values of t.

 

a.	t.10,9
b.	t.10,20
c.	t.025,82
d.	t.05,196

 

 

ANS:

 

a.	1.383
b.	1.325
c.	1.990
d.	1.653

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

176. Use the t-table to find the following probabilities.

 

a.	P(t8 > 2.306)
b.	P(t80 > 2.639)
c.	P(t24 > 1.711)
d.	P(t35 > 1.306)

 

 

ANS:

 

a.	0.025
b.	0.005
c.	0.050
d.	0.100

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

177. Use the c² table to find the following values of c².

 

a.
b.
c.
d.

 

 

ANS:

 

a.	14.61
b.	37.6
c.	4.61
d.	45.44

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

178. Use the c² table to find the following probabilities.

 

a.
b.
c.
d.

 

 

ANS:

 

a.	0.900
b.	0.050
c.	0.025
d.	0.010

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

179. Use the F table to find the following values of F.

 

a.	F.01,12,20
b.	F.05,20,40
c.	F.025,5,15
d.	F.01,8,30

 

 

ANS:

 

a.	3.23
b.	1.84
c.	3.58
d.	3.17

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

180. Use the F table to find the following values of F.

 

a.	F.99,12,20
b.	F.95,20,40
c.	F.975,5,15
d.	F.99,8,30

 

 

ANS:

 

a.	1 / 3.86 = 0.2591
b.	1 / 1.99 = 0.5025
c.	1 / 6.43 = 0.1555
d.	1 / 5.20 = 0.1923

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

181. Use the F table to find the following probabilities.

 

a.	P(F6,14 > 2.85)
b.	P(F20,60 > 2.20)
c.	P(F12,25 > 2.51)
d.	P(F15,30 > 2.01)

 

 

ANS:

 

a.	0.05
b.	0.01
c.	0.025
d.	0.05

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

182. Suppose you have a Student t distribution with 20 degrees of freedom.

 

a.	Find the mean.
b.	Find the variance.
c.	Find the standard deviation.

 

 

ANS:

 

a.	0
b.	20/18 = 1.11
c.	1.05

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

183. Suppose you have a c² distribution with 20 degrees of freedom.

 

a.	Find the mean.
b.	Find the variance.
c.	Find the standard deviation.

 

 

ANS:

 

a.	20
b.	40
c.	6.32

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

184. Suppose you have an F distribution with degrees of freedom 10, 20.

 

a.	Find the mean
b.	Find the variance.
c.	Find the standard deviation.

 

 

ANS:

 

a.	20/18 = 1.11
b.	[2(20)2 (10 + 20 – 2)] / [(10)(18)2 (20 – 4)] = 0.4321
c.	Ö0.4321 = 0.6573

 

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Application

 

185. What happens to the shape, mean, and variance of a Student t distribution as the degrees of freedom increase?

 

ANS:

As the degrees of freedom of the Student t distribution increase, the shape approaches a standard normal distribution; the mean remains 0; and the variance approaches 1.

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension

 

186. What happens to the shape, mean, and variance of a c² distribution as the degrees of freedom increase?

 

ANS:

The shape becomes less positively skewed; the mean increases to infinity, as does the variance.

 

PTS:   1                    DIF:    Easy               OBJ:   SFME.KELL.15.08.04

NAT:  BUSPROG.SFME.KELL.15.03       STA:   DISC.SFME.KELL.15.04

KEY:  Bloom’s: Comprehension